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If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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07 Jul 2013, 07:39
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If y= (x1)(x+2), then what is the least possible value of y? A. 3 B. 9/4 C. 2 D. 3/2 E. 0 Any alternative solutions?
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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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07 Jul 2013, 07:55
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fozzzy wrote: If y= (x1)(x+2), then what is the least possible value of y?
A. 3 B. 9/4 C. 2 D. 3/2 E. 0
Any alternative solutions? The function is a parabola with a positive "a" coefficient The roots are x=1 and x=2, for values between 1 and 2 it will have negative values (E is out) \(y=x^2+x2\), you can try to insert values (starting from the least) to see if it could be the answer. Example \(3=x^2+x2\) or \(x^2+x+1=0\) =>Impossible \(\frac{9}{4}= x^2+x2\) or \(x^2+x+\frac{1}{4}=0\) => \(x=0.5\) valid => CORRECT Or approach #2:Given the roots 2 and 1, the x of the vertex will be the middle point => \(x=0.5\) And the least value will be the y coordinate of the vertex (plug \(x=0.5\) into the equation).
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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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07 Jul 2013, 07:58
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fozzzy wrote: If y= (x1)(x+2), then what is the least possible value of y?
A. 3 B. 9/4 C. 2 D. 3/2 E. 0
Any alternative solutions? \(y= (x1)(x+2)=x^2+x2\). Theory:Quadratic expression \(ax^2+bx+c\) reaches its extreme values when \(x=\frac{b}{2a}\). When \(a>0\) extreme value is minimum value of \(ax^2+bx+c\) (maximum value is not limited). When \(a<0\) extreme value is maximum value of \(ax^2+bx+c\) (minimum value is not limited). You can look at this geometrically: \(y=ax^2+bx+c\) when graphed on XY plane gives parabola. When \(a>0\), the parabola opens upward and minimum value of \(ax^2+bx+c\) is ycoordinate of vertex, when \(a<0\), the parabola opens downward and maximum value of \(ax^2+bx+c\) is ycoordinate of vertex. Examples:Expression \(5x^210x+20\) reaches its minimum when \(x=\frac{b}{2a}=\frac{10}{2*5}=1\), so minimum value is \(5x^210x+20=5*1^210*1+20=15\). Expression \(5x^210x+20\) reaches its maximum when \(x=\frac{b}{2a}=\frac{10}{2*(5)}=1\), so maximum value is \(5x^210x+20=5*(1)^210*(1)+20=25\). Back to the original question:\(y= (x1)(x+2)=x^2+x2\) > y reaches its minimum (as \(a=1>0\)) when \(x=\frac{b}{2a}=\frac{1}{2}\). Therefore \(y_{min}=(\frac{1}{2}1)(\frac{1}{2}+2)=\frac{9}{4}\) Answer: B. Or:Use derivative: \(y'=2x+1\) > equate to 0: \(2x+1=0\) > \(x=\frac{1}{2}\) > \(y_{min}=(\frac{1}{2}1)(\frac{1}{2}+2)=\frac{9}{4}\) Answer: B. Hope it's clear.
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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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07 Jul 2013, 08:04
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Here's how I did this one... \(x^2 + x  2\) we can complete the square since \((x+a)^2 = x^2 + 2ax + a^2\) Here \(2ax = x\) >>\(a=\frac{1}{2}\) \((x+1/2)^2  1/4  2 = y\) subtracting 1/4 since 1/4 will be added when its squared. \((x+1/2)^2  9/4 = y\) \(y = 9/4\)
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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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07 Jul 2013, 10:41
fozzzy wrote: If y= (x1)(x+2), then what is the least possible value of y?
A. 3 B. 9/4 C. 2 D. 3/2 E. 0
Any alternative solutions? \(y = x^2+x2 \to x^2+x(2+y)=0 \to\) For real values of x, the Discriminant\((D)\geq{0} \to 1^24*1*[(2+y)]\geq{0}\to\) \(y+2\geq{\frac{1}{4} }\to y\geq{\frac{9}{4}}\) B.
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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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28 Oct 2014, 13:08
Can we also use b^2  4ac to determine the valid minimum value, as thats what I did.. I got 0 as the answer for B and I thought 0 means invalid so went for C. I want some clarification here, first if b^24ac=0, it means eq has no real soln?? Or my approach for this question was wrong ! Thanks in advance



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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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29 Oct 2014, 02:09
I did it this way The minimum (or max, if the coef is a) should lie half way between the two roots which are 2 and 1. halfway between that is 3/2. Now put in 3/2 in place of x: (3/2 1)(3/2 +2) = 9/4



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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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29 Oct 2014, 03:40
can someone tell me why it is wrong to just test all the answer values for for x and find the lowest number? That would get you to answer E (2).
I missed a question once because I assumed that a y= (x equation). Meant x equation was equal to zero. When do we assume it's a quadratic equation?



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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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29 Oct 2014, 13:57
When X has a power 2. Quad comes from the word "square"



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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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18 Mar 2016, 06:37
f(x) = (x1)(x+2) =x^2+x2
Taking derivative F'(x) = 2x+1 =0 =>x=1/2 f(x) = 1/4 1/2 2 = 9/4
Hence B



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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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04 Jul 2016, 03:58
Bunuel wrote: fozzzy wrote: If y= (x1)(x+2), then what is the least possible value of y?
A. 3 B. 9/4 C. 2 D. 3/2 E. 0
Any alternative solutions? \(y= (x1)(x+2)=x^2+x2\). Theory:Quadratic expression \(ax^2+bx+c\) reaches its extreme values when \(x=\frac{b}{2a}\). When \(a>0\) extreme value is minimum value of \(ax^2+bx+c\) (maximum value is not limited). When \(a<0\) extreme value is maximum value of \(ax^2+bx+c\) (minimum value is not limited). You can look at this geometrically: \(y=ax^2+bx+c\) when graphed on XY plane gives parabola. When \(a>0\), the parabola opens upward and minimum value of \(ax^2+bx+c\) is ycoordinate of vertex, when \(a<0\), the parabola opens downward and maximum value of \(ax^2+bx+c\) is ycoordinate of vertex. Examples:Expression \(5x^210x+20\) reaches its minimum when \(x=\frac{b}{2a}=\frac{10}{2*5}=1\), so minimum value is \(5x^210x+20=5*1^210*1+20=15\). Expression \(5x^210x+20\) reaches its maximum when \(x=\frac{b}{2a}=\frac{10}{2*(5)}=1\), so maximum value is \(5x^210x+20=5*(1)^210*(1)+20=25\). Back to the original question:\(y= (x1)(x+2)=x^2+x2\) > y reaches its minimum (as \(a=1>0\)) when \(x=\frac{b}{2a}=\frac{1}{2}\). Therefore \(y_{min}=(\frac{1}{2}1)(\frac{1}{2}+2)=\frac{9}{4}\) Answer: B. Or:Use derivative: \(y'=2x+1\) > equate to 0: \(2x+1=0\) > \(x=\frac{1}{2}\) > \(y_{min}=(\frac{1}{2}1)(\frac{1}{2}+2)=\frac{9}{4}\) Answer: B. Hope it's clear. Can we use derivatives and integration to solve such type of problems? would that a bit advanced as per the Gmat exam,however if it helps i don't see the problem in using them



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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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04 Jul 2016, 04:02



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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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04 Jul 2016, 07:03
Bunuel wrote: rhine29388 wrote: Can we use derivatives and integration to solve such type of problems? would that a bit advanced as per the Gmat exam,however if it helps i don't see the problem in using them If you know how to use any of the advanced techniques of course you can use it. Personally I've never seen a GMAT question requiring something like this. For example, I've never seen a geometry question requiring trigonometry, every GMAT geometry question can be solved without it. Always a pleasure learning from you kudos to your post



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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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05 Jul 2016, 11:52
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i did NOT do this problem like the other posters did on here...way too complicated. i just plugged in #s.
start w/ x=1. if x=1, (11)=0, so A/C= 0. x=2: (21)(4)>0, so 1 is greatest positive # we can do. ok. x=1: (2)(1) = 2. Now this is lowest #. Elim 0. x=2: (3)(0) = 0. Not quite. x=3: (4)(1) = positive. Not going in right direction. **Answer must be 2<x<0. x= 1/2: (1/22/2)(1/2+4/2)= (3/2)(3/2)= 9/4. Good! x= 3/2: (3/22/2)(3/2+4/2)= (5/2)(1/2) = 5/4. * 9/4 = 2.25; 5/4= 1.25. 9/4 = correct. A/C B = correct.



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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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08 Sep 2017, 04:46
Ans is B y = x^2 + x  2 dy/dx = 2x+1 for maxima and minima d/dx = 0 x=1/2 put x = 1/2 in y = x^2 +x 2 y = 9/4 ANSWER IS B
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Re: If y= (x1)(x+2), then what is the least possible value of y [#permalink]
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08 Sep 2017, 04:51
Ans is B Another method apart from derivation y= x^2 +x 2 lets make it a perfect square y= x^2 +x +1/4 1/4 2 y = {(x+1/2)^2 1/4} 2 y= (x+1/2)^2 9/4 therefore we have y =z^2 9/4 z can be minimum 0 when x=1/2 so y will be minimum put z=0 so y=9/4
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